Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Integral question - $$\int\frac{\sqrt{\tan(x)}}{{\cos^2(x)}}dx$$
I see that $\frac{1}{\cos^2(x)}$ is the derivative of $\tan(x)$ so I set $t = \tan(x)$? or the whole square?

share|cite|improve this question
$\tan'x=\sec^2x$ – 1015 May 1 '13 at 20:51
Just $\tan(x)$... – David Mitra May 1 '13 at 20:52
up vote 3 down vote accepted

It is easier to set just $\tan(x) = t$. If you set $\tan(x) = t$, we get that $\sec^2(x) dx = dt$, i.e., $\dfrac{dx}{\cos^2(x)} = dt$. Hence, we get $$\int \dfrac{\sqrt{\tan(x)}}{\cos^2(x)}dx = \int \sqrt{t} dt$$ I trust you can take it from here.

share|cite|improve this answer

Hint: for a derivable function $\,f(x)\,$ , we have that

$$\int\sqrt{f(x)}\;f'(x)\;dx=\frac23f(x)\sqrt{f(x)} + C$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.